Rivest-Vuillemin Conjecture Is True for Monotone Boolean Functions with Twelve Variables. Technical Report

Transcription

1 Rivest-Vuillemin Conjecture Is True or Monotone Boolean Functions with Twelve Variables Technical Report Department o Computer Science and Engineering University o Minnesota EECS Building 200 Union Street SE Minneapolis, MN USA TR Rivest-Vuillemin Conjecture Is True or Monotone Boolean Functions with Twelve Variables Sui-xiang Gao, Ding-zhu Du, Xiao-dong Hu, and Xiaohua Jia October 02, 2000

2

3 Rivest-Vuillemin Conjecture Is True or Monotone Boolean Functions with Twelve Variables Sui-Xiang Gao x Ding-Zhu Du yzx Xiao-Dong Hu zx Xiaohua Jia x Abstract A Boolean unction (x1; x2; ; xn) is elusive i every decision tree computing must examine all n variables in the worst case. It is a long-standing conjecture that every nontrivial monotone weakly symmetric Boolean unction is elusive. In this paper, we prove this conjecture or Boolean unctions with twelve variables. Keywords. Monotone Boolean unction, decision tree, elusive. AMS(MOS) subject classications. 05C25, 68Q05, 68R05 Department o Mathematics, Graduate School at Beijing, University o Science and Technology o China, Beijing , China. y Department o Computer Science and Engineering, University o Minnesota, Minneapolis, MN 55455, USA. Supported in part by NSF under grant CCR z Institute o Applied Mathematics, Chinese Academy o Sciences, Beijing , China. Supported in part by 973 Inormation Technology and High-Perormance Sotware Program o China. x Department o Computer Science, City University o Hong Kong, Kowloon Tong, Hong Kong. 1

4 1 Introduction An assignment or a Boolean unction o n variables can be considered as a binary string o length n, i.e., a string in 0; 1g n. An assignment x o Boolean unction (x) is called a truthassignment i (x) = 1, and alse-assignment i (x) = 0. We denote by truth(x) and alse(x) respectively sets o variables with value 1 and with value 0 in the assignment x. For two assignments o a Boolean unction (x 1 ; x 2 ; ; x n ), say, x = (x 1 ; x 2 ; ; x n ) and y = (y 1 ; y 2 ; ; y n ), i x i y i or all i, then we write x y. A Boolean unction (x) is increasing i (x) = 1 and x y imply (y) = 1, and decreasing i (y) = 1 and x y imply (x) = 1, monotone i it is either increasing or decreasing. (x) is nontrivial i it is not a constant unction. A decision tree o a Boolean unction is a rooted binary tree, whose nonlea vertices are labeled by its variables, and leaves are labeled by 0 and 1. Edges o this binary tree are also labeled by 0 and 1 such that edges rom an non-lea vertex to its two children are labeled by 0 and 1 respectively, and every variable appears at most once in a path rom the root to a lea. Given an assignment to variables o a Boolean unction (x 1 ; x 2 ; ; x n ), we can compute the unction value o by its decision tree as ollows: starting rom the root, we look at its label. I its label is x i, then we make a decision according to the value o x i, to decide where we go. I x i = 0, then we go to the next vertex along the edge with label 0; i x i = 1, then we go to the next vertex along the edge with label 1. Once a lea is reached, the unction value or the given assignment is obtained rom the label o the lea. Each decision tree o gives an algorithm to compute the unction value. The computation time depends on the length o root-lea path that is the number o variables on the path. The depth o a decision tree is the maximum length o all paths rom the root to leaves. A Boolean unction generally has many decision trees. We denote by D() the minimum depth o all decision trees computing Boolean unction. D() is called the decision tree complexity o. Clearly, D() n i has n variables. (x 1 ; ; x n ) is said to be elusive i D() = n. The decision tree complexity is closely related to several other combinatorial and complexity issues, such as the certicate complexity (see [1]), the block sensitivity ([2]), the packing o graphs (see [3]), and the time-complexity o a CREW PRAM (see [2]). A group G o permutations on 1; 2; ; ng is called transitive i or any i; j 2 1; 2; ; ng, there exists 2 G such that (i) = j. Let (x 1 ; x 2 ; ; x n ) be a Boolean unction and G be a group o permutations on 1; 2; ; ng. (x 1 ; x 2 ; ; x n ) is said to be invariant under group G i or any 2 G, (x 1 ; x 2 ; ; x n ) = (x (1) ; x (2) ; ; x (n) ): A Boolean unction (x 1 ; x 2 ; ; x n ) is said to be weakly symmetric i there exists a transitive permutation group G on 1; 2; ; ng such that (x 1 ; x 2 ; ; x n ) is invariant under G. There is an interesting conjecture on monotone weakly symmetric Boolean unctions. 2

5 Rivest-Vuillemin Conjecture (1975): Any nontrivial monotone weakly symmetric Boolean unction (x 1 ; x 2 ; ; x n ) is elusive. Rivest and Vuillemin [11] proved that this conjecture is true when n is a prime power. 1 Gao et al [6, 7] showed that Rivest-Vuillemin conjecture is true or n = 6; 10. In this paper, we show that Rivest-Vuillemin Conjecture is true or n = 12. The proo involves some new techniques developed based on some acts on permutation groups. 2 Preliminary An abstract complex on a nite set X is a amily o subsets o X, such that i A is a member o, so is every subset o A. Each member o is called a ace o. A maximal ace o abstract complex is a ace that is not contained by another ace. A ree ace is a non-maximal ace that is contained by only one maximal ace. An elementary collapse is an operation that deletes a ree ace together with all aces containing it. An abstract complex is collapsible i it can be collapsed to the empty abstract complex through a sequence o elementary collapses. by The complex o a monotone Boolean unction (x 1 ; x 2 ; ; x n ) is an abstract complex dened =( alse(x) j (x) = 1g; i is increasing truth(x) j (x) = 1g; i is decreasing: Each vertex o is a variable o. The ollowing can be ound in [9]. Lemma 2.1 Let be a nontrivial monotone Boolean unction. I is not elusive, then collapsible. is For an abelian group G, an abstract complex is G-acyclic i the homology groups o under G are H 0 (; G) = G; i = 0; H i (; G) = 0; i > 0: The ollowing can be ound in [9]. Lemma 2.2 I is collapsible, then is Z p -acyclic. The ollowing ollows immediately rom Lemma 2.1 and Lemma 2.2. Corollary 2.1 Let be a nontrivial monotone Boolean unction. I is not elusive, then is Z p -acyclic. 1 Actually, they proved that i (0; ; 0) 6= (1; ; 1) and n is a prime power, then weakly symmetric Boolean unction (x1; ; xn) is elusive. They also made a conjecture or general n with condition (0; ; 0) 6= (1; ; 1) instead o monotoneness. Illies in 1978 ound a counterexample or Rivest-Vuillemin's original conjecture (see [5]). Current Rivest-Vuillemin conjecture is a modication suggested by this counterexample. 3

6 X The Euler characteristic o an abstract complex is dened by X () = A2;A6=;(?1) jaj?1 = (?1) jaj?1 + 1; A2 in particular, (;g) = 0 and dene (;) = 1. A permutation on the vertex set o an abstract complex is called an automorphism o, i or each ace A 2, (A) = (a) j a 2 Ag is still a ace o. Every invariant permutation o Boolean unction induces an automorphism o. Let G be a group o automorphisms on, an orbit o G is a minimal subset o vertices o which is closed under actions o G. Clearly, G has only one orbit on i and only i G is transitive on vertices o. Dene G = H 1 ; ; H k g j H 1 ; ; H k are orbits o G, and H 1 [ [ H k 2 g [ ;g: G is an abstract complex (see [5]). For p and q primes, denote by Y q p the class o nite groups G with normal subgroups P /H /G, such that P is o p-power order, the quotient group G=H is o q-power order, and the quotient group H=P is cyclic; denote by Y p the class o nite groups G with normal p-subgroups P / G such that the quotient group G=P is cyclic. The ollowing lemma comes rom [10]. Lemma 2.3 Let G be a group o automorphisms on a collapsible abstract complex. (1) I G is a cyclic group or G 2 Y p or some prime p, then ( G ) = 1. (2) I G 2 Yp q or some primes p and q, then (G ) 1 (mod q). The ollowing lemma ollows rom previous ones. Lemma 2.4 I a nontrivial monotone Boolean unction has a transitive cyclic invariant group or has a transitive invariant group in Y p or in Yp q, then is elusive. Proo. Let G be a group that meets the conditions o the current lemma, then G has only one orbit in since it is transitive. This orbit cannot be a ace o. In act, i it is a ace, then the monotonicity o orces that must be a constant, contradicting the hypothesis that is nontrivial. Thus, G = ;g and (G ) = 0. By Lemma 2.3, is not collapsible. By Lemma 2.1, is elusive. 2 This lemma will be used extensively together with many acts on group theory in the next section, such as acts about block systems. For a transitive permutation group G on a set, a partition ( 1 ; ; k ) o is called a block system o G i each i is transormed to some j under the action o any element o G. G is primitive i G has no nontrivial block system; otherwise, G is imprimitive. 4

7 For simplicity, all involved terminologies and concepts on group theory are employed rom the same reerence [4] while results come rom dierent sources. 3 Main Result In this section, we prove the ollowing main result. Main Theorem. Every nontrivial monotone weakly symmetric Boolean unction (x 1 ; x 2 ; ; x n ) is elusive when n = 12. To prove it, we rst show some lemmas. Lemma 3.1 [13] Any transitive permutation group o twelve degree contains one o ollowing minimal transitive groups as its subgroup: T 1 = h(1; 3; 5; 7; 9; 11; 2; 4; 6; 8; 10; 12)i; T 2 = h(1; 6; 9; 2; 5; 10)(3; 8; 11; 4; 7; 12); (1; 7)(2; 8)(3; 9)(4; 10)(5; 11)(6; 12)i; T 3 = h(1; 10; 5; 2; 9; 6)(3; 12; 7; 4; 11; 8); (1; 3)(2; 4)(5; 11)(6; 12)(7; 9)(8; 10)i; T 4 = h(1; 6; 9; 2; 5; 10)(3; 8; 11; 4; 7; 12); (1; 3; 2; 4)(5; 11; 6; 12)(7; 10; 8; 9)i; T 5 = h(1; 9; 5)(2; 10; 6)(3; 11; 7)(4; 12; 8); (1; 12; 7)(2; 11; 8)(3; 10; 5)(4; 9; 6)i; T 6 = h(1; 8; 11)(2; 7; 12)(3; 5; 10)(4; 6; 9); (1; 9; 7; 4; 11; 5)(2; 10; 8; 3; 12; 6)i; T 7 = h(1; 4)(2; 3)(5; 12)(6; 11)(7; 10)(8; 9); (1; 11; 7)(2; 12; 8)(3; 10; 6)(4; 9; 5)i; T 8 = h(1; 8; 4; 10)(2; 7; 3; 9)(5; 12; 6; 11); (1; 3; 6)(2; 4; 5)(7; 12; 10)(8; 11; 9)i; T 9 = h(1; 2)(3; 4)(5; 7; 6; 8)(9; 11; 10; 12); (1; 11; 6)(2; 12; 5)(3; 10; 8)(4; 9; 7)i; T 10 = h(1; 7)(2; 8)(3; 9; 5; 11)(4; 10; 6; 12); (1; 6; 3; 2; 5; 4)(7; 8)(9; 10)(11; 12)i; T 11 = h(1; 9)(2; 10)(3; 11)(4; 12)(5; 7)(6; 8); (1; 6; 3; 2; 5; 4)(7; 10)(8; 9)(11; 12)i; T 12 = h(1; 11; 3; 10)(2; 12)(4; 7)(5; 8; 6; 9); (1; 4; 10; 7; 2; 6; 11; 9)(3; 5; 12; 8)i; T 13 = h(1; 5; 3; 4)(2; 6)(7; 12; 9; 11)(8; 10); (1; 8)(2; 9; 3; 7)(4; 11)(5; 10; 6; 12)i; T 14 = h(1; 5; 12; 2; 6; 11)(3; 8; 10; 4; 7; 9); (1; 7; 11; 2; 8; 12)(3; 6; 10; 4; 5; 9)i; T 15 = h(1; 9; 6; 12; 2; 10; 5; 11)(3; 8; 4; 7); (1; 2)(5; 6)(7; 12; 10)(8; 11; 9)i; T 16 = h(1; 12; 5; 3; 11; 6; 2; 10; 7)(4; 9; 8); (1; 3; 2)(5; 8; 6)(9; 11; 12)i; T 17 = h(1; 7)(2; 9; 3; 8)(4; 11; 6; 10; 5; 12); (1; 4; 2; 6; 3; 5)(7; 12; 8; 11)(9; 10)i; where h; i denotes the group generated by permutations and. imprimitive groups and their orders are as ollows: All o these groups are group : T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8 T 9 T 10 T 11 T 12 T 13 T 14 T 15 T 16 T 17 order : Lemma 3.2 T 2 ; T 3 2 Y 2. Proo. For T 2, consider H = h(1; 7)(2; 8)(3; 9)(4; 10)(5; 11)(6; 12)i = (1); (1; 7)(2; 8)(3; 9)(4; 10)(5; 11)(6; 12)g: 5

8 Note that H is a normal subgroup o T 2 and the quotient group T 2 =H = h(1; 6; 9; 2; 5; 10)(3; 8; 11; 4; 7; 12)i: Thus, T 2 2 Y 2. T 3 2 Y 2 can be shown similarly. 2 Lemma 3.3 T 4 2 Y 3. Proo. Denote a = (1; 6; 9; 2; 5; 10)(3; 8; 11; 4; 7; 12) b = (1; 3; 2; 4)(5; 11; 6; 12)(7; 10; 8; 9): Then T 4 = ha; bi = (1); a; a 2 ; a 3 ; a 4 ; a 5 ; b; b 3 ; ab; ba; a?1 b?1 ; b?1 a?1 g: Consider H = ha 2 i = (1); a 2 ; a?2 g: It is easy to veriy that (1) H is a normal subgroup o T 4, (2) jhj = 3, and (3) T 4 =H = hbi. Thereore, T 4 2 Y 3. 2 Lemma 3.4 T 5 2 Y 2. Proo. Denote a = (1; 9; 5)(2; 10; 6)(3; 11; 7)(4; 12; 8) b = (1; 12; 7)(2; 11; 8)(3; 10; 5)(4; 9; 6): Then T 5 = ha; bi = (1); a; b; a 2 ; b 2 ; ab; ba; a 2 b; ab 2 ; a 2 b 2 ; b 2 a 2 ; a 2 bab 2 g: Consider H = (1); a 2 bab 2 ; a 2 b; ab 2 g: Note that a?1 = a 2, b?1 = b, a 2 b = b 2 a and ab 2 = ba 2. It is easy to veriy the ollowing: (1) H is a normal subgroup o T 5. (2) jhj = 4: (3) T 5 =H = hai. Thus, T 5 2 Y

9 Lemma 3.5 T 6 2 Y 2. Proo. Denote Then a = (1; 8; 11)(2; 7; 12)(3; 5; 10)(4; 6; 9) b = (1; 9; 7; 4; 11; 5)(2; 10; 8; 3; 12; 6): T 6 = ha; bi = (1); a; a 2 ; b; b 2 ; b 3 ; b 4 ; b 5 ; ab; ab 2 ; ab 3 ; ab 4 ; ab 5 ; a 2 b; a 2 b 2 ; a 2 b 3 ; a 2 b 4 ; ba; ba 2 ; b 2 a 2 ; b 4 a; b 4 a 2 ; b 5 a; ab 2 a; ab 5 ag and the ollowing equalities can be derived by simple computations: Consider a?1 = a; b?1 = b 5 ; ab 3 = b 3 a; ab 4 = b 2 a 2 ; b 4 a = a 2 b 2 ; a 2 b 3 = b 3 a 2 ; a 2 b 5 = ba; b 5 a 2 = ab; (b 2 a 2 )(a 2 b 2 ) = (a 2 b 2 )(b 2 a 2 ) = (ab)(ba): H = (1); (1; 2)(3; 4)(5; 6)(7; 8); (1; 2)(3; 4)(9; 10)(11; 12); (5; 6)(7; 8)(9; 10)(11; 12)g = (1); a 2 b 2 ; b 2 a 2 ; (a 2 b 2 )(b 2 a 2 )g The ollowing can be veried easily by using above equalities: (1) H is a normal subgroup o T 6 ; (2) jhj = 4; (3) T 6 =H = hbi: Thereore, T 6 2 Y 2. 2 Lemma 3.6 T 7 2 Y 2 2 Proo. Denote Then a = (1; 4)(2; 3)(5; 12)(6; 11)(7; 10)(8; 9) b = (1; 11; 7)(2; 12; 8)(3; 10; 6)(4; 9; 5): T 7 = ha; bi = (1); a; b; b 2 ; ab; ab 2 ; ba; b 2 a; (ab) 2 ; (ba) 2 ; (ab)(ba); aba; bab; b 2 ab; bab 2 ; b 2 ab 2 ; b 2 aba; abab 2 ; ab 2 ab; bab 2 a; ab 2 aba; abab 2 a; (ab) 2 (ba) 2 ; a(ab) 2 (ba) 2 g and the ollowing equalities can be derived by simple computations: a?1 = a; b?1 = b 2 ; (ab)?1 = b 2 a; (ba)?1 = ab 2 ; (ba)(ab) = b 2 ; ab = (b 2 a) 3 ; (ab) 2 = (b 2 a) 2 ; (ab) 3 = b 2 a; ba = (ab 2 ) 3 ; (ba) 2 = (ab 2 ) 2 ; (ba) 3 = ab 2 ; babab = ab 2 a; (bab) 2 = (ab) 2 (ba) 2 = a(bab) 2 a: 7

10 Let and H = hb; abai = (1); b; b 2 ; aba; (aba) 2 ; (aba)b; b(aba); b 2 (aba); (aba)b 2 ; b(aba) 2 ; (aba) 2 b; (aba)b 2 (aba)g = (1); (1; 2)(3; 4)(5; 6)(7; 8); (1; 2)(3; 4)(9; 10)(11; 12); (5; 6)(7; 8)(9; 10)(11; 12); (1; 11; 7)(2; 12; 8)(3; 10; 6)(4; 9; 5); (1; 7; 11)(2; 8; 12)(3; 6; 10)(4; 5; 9); (1; 11; 8)(2; 12; 7)(3; 10; 5)(4; 9; 6); (1; 8; 11)(2; 7; 12)(3; 5; 10)(4; 6; 9); (1; 12; 7)(2; 11; 8)(3; 9; 6)(4; 10; 5); (1; 7; 12)(2; 8; 12)(3; 6; 9)(4; 5; 10); (1; 12; 8)(2; 11; 7)(3; 9; 5)(4; 10; 6); (1; 8; 12)(2; 7; 11)(3; 5; 9)(4; 6; 10)g P = (1); abab; baba; (abab)(baba)g = (1); (1; 2)(3; 4)(5; 6)(7; 8); (1; 2)(3; 4)(9; 10)(11; 12); (5; 6)(7; 8)(9; 10)(11; 12)g: Using above equalities, we can veriy the ollowing : (1) H is a normal subgroup o T 7 and P is a normal subgroup o H; (2) jp j = 4; (3) jt 7 =Hj = 2; (4) H=P = hb 2 i. Thus, T 7 2 Y Lemma 3.7 T 8 2 Y 3. Proo. Denote Then a = (1; 8; 4; 10)(2; 7; 3; 9)(5; 12; 6; 11) b = (1; 3; 6)(2; 4; 5)(7; 12; 10)(8; 11; 9): T 8 = ha; bi = (1); a; a 2 ; a 3 ; b; b 2 ; ab; a 2 b; a 3 b; ab 2 ; a 2 b 2 ; a 3 b 2 ; ba; ba 3 ; b 2 a; b 2 a 3 ; bab; bab 2 ; ba 3 b; b 2 ab 2 ; b 2 ab; b 2 a 3 b; (ab) 2 ; (a 3 b) 2 ; (ab 2 ) 2 ; (a 3 b 2 ) 2 ; (a 2 b 2 )(ab) 2 ; (ab)(a 3 b); (a 3 b)(ab); a 3 b 2 ab 2 ; a 3 ba; aba 3 ; ab 2 a; aba; ba 3 b 2 ; a 2 bab 2 g; and the ollowing equalities can be derived by simple computations: a?1 = a 3 ; b?1 = b 2 ; a 2 = ba 2 b = b 2 a 2 b 2 ; b = a 2 b 2 a 2 ; b 2 = a 2 ba 2 ; a 2 b 2 = ba 2 ; Let b 2 a 2 = a 2 b; (ab)?1 = b 2 a 3 ; (ba)?1 = a 3 b 2 ; (ab 2 )?1 = ba 3 ; (b 2 a)?1 = a 3 b: H = (1); b; b 2 ; a 3 ba; aba 3 ; aba 3 b 3 ; a 3 bab; aba 3 b; a 3 bab 2 g = (1); (1; 3; 6)(2; 4; 5)(7; 12; 10)(8; 11; 9); (1; 6; 3)(2; 5; 4)(7; 10; 12)(8; 9; 11); (1; 3; 6)(2; 4; 5)(7; 10; 12)(8; 9; 11); (1; 6; 3)(2; 5; 4)(7; 12; 10)(8; 11; 9); (1; 3; 6)(2; 4; 5); (1; 6; 3)(2; 5; 4); (7; 10; 12)(8; 9; 11); (7; 12; 10)(8; 11; 9)g: 8

11 Then one can veriy the ollowing with above equalities: (1) H is a normal subgroup o T 8 ; (2) jhj = 9; (3) T 8 =H = hai. Thus, T 8 2 Y 3. 2 The ollowing two lemmas can be ound in [14]. Lemma 3.8 Let G be an imprimitive group. Suppose 1, 2,, s orm a block system o G. Each element g o G induces a permutation on this block system: g = 1 2 s! g 1 g : 2 g s Denote Q = g j g 2 Gg. Then Q is a group o permutations o degree s. Lemma 3.9 Let G be an imprimitive group. Suppose 1 ; 2 ; ; s orm a block system o G. Assume that Q is dened in Lemma 3.8 and H = g 2 G j g i = i; i = 1; 2; ; sg. Then H / G and G=H = Q. Lemma 3.10 T 9 2 Y 2. Proo. Let Then T 9 = ha; bi. It can be easily veried that a = (1; 2)(3; 4)(5; 7; 6; 8)(9; 11; 10; 12) b = (1; 11; 6)(2; 12; 5)(3; 10; 8)(4; 9; 7): 1 = 1; 2; 3; 4g; 2 = 5; 6; 7; 8g; 3 = 9; 10; 11; 12g orm a block system o T 9. The generators a and b o T 9 induce respectively a permutation on 1 ; 2 ; 3 g: a : 1! 1 ; 2! 2 ; 3! 3 : b : 1! 3 ; 2! 1 ; 3! 2 : Obviously, a = (1) and b = (1; 3; 2). Since every permutation g in T 9 can be written as a product o powers o a and b, the induced permutation g on 1 ; 2 ; 3 g can be written as a product o powers o a and b. Thereore, Q = g jg 2 T 9 g = h a ; b i = h b i: By Lemma 3.1, T 9 is an imprimitive group. By Lemma 3.9, there exists a subgroup H such that H / T 9 and T 9 =H = Q. jhj = jt 9 j=jqj = 2 4 since jt 9 j = 48 and jqj = 3. Moreover, Q is a cyclic group. Thus, T 9 2 Y 2. 2 The ollowing two lemmas can be ound in [12]. 9

12 Lemma 3.11 Any group o prime order is cyclic. Lemma 3.12 Any subgroup with index 2 is a normal subgroup. Lemma 3.13 T 10 ; T 11 ; T 12 ; T 13 2 Y 2 3. Let a 10 = (1; 7)(2; 8)(3; 9; 5; 11)(4; 10; 6; 12) b 10 = (1; 6; 3; 2; 5; 4)(7; 8)(9; 10)(11; 12) a 11 = (1; 9)(2; 10)(3; 11)(4; 12)(5; 7)(6; 8) b 11 = (1; 6; 3; 2; 5; 4)(7; 10)(8; 9)(11; 12) a 12 = (1; 11; 3; 10)(2; 12)(4; 7)(5; 8; 6; 9) b 12 = (1; 4; 10; 7; 2; 6; 11; 9)(3; 5; 12; 8) a 13 = (1; 5; 3; 4)(2; 6)(7; 12; 9; 11)(8; 10) b 13 = (1; 8)(2; 9; 3; 7)(4; 11)(5; 10; 6; 12): Then T 10 = ha 10 ; b 10 i, T 11 = ha 11 ; b 11 i, T 12 = ha 12 ; b 12 i, and T 13 = ha 13 ; b 13 i. It is easy to veriy that 1 = 1; 3; 5g; 2 = 2; 4; 6g; 3 = 7; 9; 11g; 4 = 8; 10; 12g orm a block system o T 10 and T 11, and 0 1 = 1; 2; 3g; 0 2 = 4; 5; 6g; 0 3 = 7; 8; 9g; 0 4 = 10; 11; 12g orm a block system o T 12 and T 13. The generators a 10 and b 10 o T 10 induce ollowing permutations on 1 ; 2 ; 3 ; 4 g as ollows: a10 : 1! 3 ; 2! 4 ; 3! 1 ; 4! 2 : b10 : 1! 2 ; 2! 1 ; 3! 4 ; 4! 3 : That is, a10 = (1; 3)(2; 4) and b10 = (1; 2)(3; 4). The generators a 11 and b 11 o T 11 induce ollowing permutations on 1 ; 2 ; 3 ; 4 g: a11 : 1! 3 ; 2! 4 ; 3! 1 ; 4! 2 : b11 : 1! 2 ; 2! 1 ; 3! 4 ; 4! 3 : That is, a11 = (1; 3)(2; 4) and b11 = (1; 2)(3; 4). The generators a 12 and b 12 o T 12 induce the ollowing permutations on 0 1 ; 0 2 ; 0 3 ; 0 4 g: a12 : 0 1! 0 4; 0 2! 0 3; 0 3! 0 2; 0 4! 0 1: 10

13 That is, a12 = (1; 4)(2; 3) and b12 = (1; 2; 4; 3). b12 : 0 1! 0 2 ; 0 2! 0 4 ; 0 3! 0 1 ; 0 4! 0 3 : The generators a 13 and b 13 o T 13 induce the ollowing permutations on 0 1 ; 0 2 ; 0 3 ; 0 4 g: a13 : 0 1! 0 2; 0 2! 0 1; 0 3! 0 4; 0 4! 0 3: b13 : 0 1! 0 3; 0 2! 0 4; 0 3! 0 1; 0 4! 0 2: That is, a13 = (1; 2)(3; 4) and b13 = (1; 3)(2; 4). Since every permutation g in T i can be generated by a i and b i, the induced permutation g can be generated by ai and bi ( i = 10; 11; 12; 13 ). Thus, Q 10 = g j g 2 T 10 g = h a10 ; b10 i = (1); (1; 2)(3; 4); (1; 3)(2; 4); (1; 4)(2; 3)g; Q 11 = g j g 2 T 11 g = h a11 ; b11 i = (1); (1; 3)(2; 4); (1; 2)(3; 4); (1; 4)(2; 3)g; Q 12 = g j g 2 T 12 g = h a12 ; b12 i = (1); (1; 4)(2; 3); (1; 2; 4; 3); (1; 3; 2; 4)g; Q 13 = g j g 2 T 13 g = h a13 ; b13 i = (1); (1; 2)(3; 4); (1; 3)(2; 4); (1; 4)(2; 3)g: All Q i or i = 10; 11; 12; 13 are groups o order 4. For i = 10; 11; 12; 13, T i is imprimitive by Lemma 3.1. By Lemma 3.9, there exists an normal subgroup H i / T i such that T i =H i = Qi. Moreover, the acts that jt i j = 72 and jq i j = 4 imply that jh i j = 18. Since 18 = 2 3 2, H i has a Sylow subgroup K i o order 3 2. By Lemma 3.12, K i is a normal subgroup o H i. Since jh i =K i j = 2, H i =K i is a cyclic group by Lemma Thereore, T i 2 Y 2 3 or i = 10; 11; 12; Lemma 3.14 T 14 2 Y 2. Proo. Let Then T 14 = ha; bi. It is easy to veriy that a = (1; 5; 12; 2; 6; 11)(3; 8; 10; 4; 7; 9) b = (1; 7; 11; 2; 8; 12)(3; 6; 10; 4; 5; 9) 1 = 1; 2; 3; 4g; 2 = 5; 6; 7; 8g; 3 = 9; 10; 11; 12g orm a block system o T 14. The generators a and b o T 14 induce ollowing permutations on 1 ; 2 ; 3 g: a : 1! 2 ; 2! 3 ; 3! 1 : b : 1! 2 ; 2! 3 ; 3! 1 : That is, a = b = (1; 2; 3). Since every permutation g o T 14 can be generated by a and b, the induced permutation g can be generated by a and b. Q = g j g 2 T 14 g = h a ; b i = h a i; 11

14 which is a cyclic group o order 3. Since T 14 is imprimitive, by Lemma 3.9 there exists an normal subgroup H / T 14 such that T 14 =H = Q. Moreover, jt 14 j = 96 and jqj = 3 imply that jhj = 2 5. Thereore, T 14 2 Y 2. 2 The ollowing two lemmas can be ound in [10]. Lemma 3.15 I a p-group P acts on the nite complex, then ( P ) () i is Z p -acyclic, so is P : (mod p), and Lemma 3.16 Suppose Z n acts on the nite Q-acyclic complex. Then ( Zn ) = 1: Lemma 3.17 Let (x 1 ; x 2 ; ; x 12 ) be a nontrivial monotone Boolean unction, invariant under the action o T 15. I is not elusive, then ( T 15 ) 1 (mod 3). Proo. Let Then T 15 = ha; bi. It is easy to veriy that a = (1; 9; 6; 12; 2; 10; 5; 11)(3; 8; 4; 7) b = (1; 2)(5; 6)(7; 12; 10)(8; 11; 9): 1 = 1; 2; g; 2 = 3; 4g; 3 = 5; 6g; 4 = 7; 8g; 5 = 9; 10g; 6 = 11; 12g orm a block system o T 15. The generators a and b o T 15 induce ollowing permutations on 1 ; 2 ; 3 ; 4 ; 5 ; 6 g: a : 1! 5 ; 2! 4 ; 3! 6 ; 4! 2 ; 5! 3 ; 6! 1 : b : 1! 1 ; 2! 2 ; 3! 3 ; 4! 6 ; 5! 4 ; 6! 5 : That is, a = (1; 5; 3; 6)(2; 4) and b = (4; 6; 5). Since every permutation g o T 15 can be generated by a and b, g can be generated by a and b. Hence, Q = g jg 2 T 15 g = h a ; b i = (1); (12)(45); (12)(46); (12)(56); (13)(45); (13)(46); (13)(56); (23)(45); (23)(46); (23)(56); (123); (132); (456); (465); (1; 2; 3)(456); (123)(465); (132)(456); (132)(465); (1425)(36); (1524)(36); (1426)(35); (1624)(35); (1435)(26); (1534)(26); (1436)(25); (1634)(25); (1526)(34); (1625)(34); (1536)(24); (1635)(24); (14)(2536); (14)(2635); (15)(2436); (15)(2634); (16)(2435); (16)(2534)g; which is a group o order 36. Since T 15 is imprimitive, by Lemma 3.9 there exists a normal subgroup H / T 15 such that T 15 =H = Q. Moreover, jt 15 j = 576 and jqj = 36 imply jhj = 2 4. Let K = h(123); (456)i = (1); (123); (132); (456); (123)(456); (123)(465); (132)(456); (132)(465)g: 12

15 Then K / Q, jkj = 9, jq=kj = 4, and Q=K = h(1536)(24)i. I is not elusive, then is Z 2 -acyclic by Corollary 2.1. Since H and Q=K are o 2-power order, H and thus ( H ) Q=K are Z 2 -acyclic by Lemma By Lemmas 3.15 and 3.16, we have ( T 15 ) = (( H ) T15=H ) = (( H ) Q ) = ((( H ) Q=K ) K ) (( H ) Q=K ) = 1 (mod 3): The next lemma can be ound in [8]. 2 Lemma 3.18 Suppose that G is a group o order pq t, where p and q are primes, then G has either a normal subgroup o order q t or a normal subgroup o order pq t?1 : Lemma 3.19 Let (x 1 ; x 2 ; ; x 12 ) be a nontrivial monotone Boolean unction, invariant under the action o T 16. I is not elusive, then ( T 16 ) 1 (mod 3). Proo. Let Then T 16 = ha; bi. It is easy to veriy that a = (1; 12; 5; 3; 11; 6; 2; 10; 7)(4; 9; 8) b = (1; 3; 2)(5; 8; 6)(9; 11; 12): 1 = 1; 2; 3; 4g; 2 = 5; 6; 7; 8g; 3 = 9; 10; 11; 12g orm a block system o T 16. The generators a and b o T 16 induce ollowing permutations on 1 ; 2 ; 3 g: a : 1! 3 ; 2! 1 ; 3! 2 : b : 1! 1 ; 2! 2 ; 3! 3 : That is, a = (1; 2; 3) and b = (1). Since every permutation g o T 16 can be generated by a and b, g can be generated by a and b. Thus, which is a cyclic group o order 3. Q = g jg 2 T 16 g = h a ; b i = h a i; Since T 16 is imprimitive, by Lemma 3.9 there exists a normal subgroup H 1 / T 16 such that T 16 =H 1 = Q. Moreover, jt16 j = 576 and jqj = 3 imply that jh 1 j = 192 = By Lemma 3.18, there exists a normal subgroup H 2 o H 1 such that either jh 2 j = 2 6 and jh 1 =H 2 j = 3 or jh 2 j = and jh 1 =H 2 j = 2. In the ormer case, T 16 2 Y2 3, and by Lemmas 2.1 and 2.3, ( T 16 ) 1 (mod 3). In the latter case, applying Lemma 3.19 to H 2, we can nd there exists a normal subgroup H 3 o H 2 such that either jh 3 j = 2 5 and jh 2 =H 3 j = 3 or jh 3 j = and jh 2 =H 3 j = 2. Repeating this process, we can nally reach one o the ollowing cases: 13

16 (1) For some integer k with 7 > k > 2, there exists a normal subgroup chain such that T 16. H 1. H 2.. H k?1. H k jt 16 =H 1 j = 3; jh i =H i+1 j = 2(i = 1; 2; ; k? 2); jh k?1 =H k j = 3 and jh k j = 2 8?k (2) there exists a normal subgroup chain T 16. H 1. H 2. H 3. H 4. H 5. H 6. H 7 such that jt 16 =H 1 j = 3, jh i =H i+1 j = 2 (i = 1; 2; ; 6) and jh 7 j = 3: In the ormer case, ( T 16 ) = (( H 1 )T 16=H 1 ) ( H 1 ) (mod 3) by Lemma Moreover, since is not elusive, is Z 2 -acyclic by Corollary 2.1. Denote Then 1 is Z 2 -acyclic by Lemma 3.15, and 1 = (( H 1=H 2 ) H 2=H 3 ) H k?2=hk?1 : ( H 1 ) = ((((H 1=H 2 ) H 2=H 3 ) H k?2=hk?1 ) H k?1 ) = ( H k?1 1 ): Furthermore, ( H k?1 ) = 1 by Lemma 2.3 since H k is a normal 2-subgroup o H k?1 and H k?1 =H k is a cyclic group o order 3. Thus, ( T 16 ) 1 (mod 3). In the latter case, ( T 16 ) = (( H 1 )T 16=H 1 ) ( H 1 ) (mod 3) by Lemma I is not elusive, then is Z 2 -acyclic by Corollary 2.1. Denote 1 = ((((( H 1=H 2 ) H 2=H 3 ) H 3=H 4 ) H 4=H 5 ) H 5=H 6 ) H 6=H 7 : Then 1 is Z 2 -acyclic by Lemma 3.15 since jh i =H i+1 j = 2 (i = 1; 2; ; 6). Thus, ( H 1 ) = (((((((H 1=H 2 ) H 2=H 3 ) H 3=H 4 ) H 4=H 5 ) H 5=H 6 ) H 6=H 7 ) H 7 ) = ( H 7 1 ): Moreover, ( H 7 1 ) = 1 by Lemma 3.16 since H 7 is cyclic. This implies ( T 16 ) 1 (mod 3). 2 Lemma 3.20 Suppose G acts on the nite Z p -acyclic complex. I there exists a sequence o subgroups P / H / G such that (1) G=H is o q-power order, (2) H=P is o p-power order, and (3) P is cyclic, then ( G ) 1 (mod q). 14

17 Proo. Since is Z p -acyclic and H=P is o p-power order, H=P is Z p -acyclic by Lemma Denote 1 = H=P, then 1 is Z p -acyclic and hence Q-acyclic. By Lemma 3.16, ( P 1 ) = 1. This, combined with Lemma 3.15, shows that ( G ) = (( H ) G=H ) ( H ) = (( H=P ) P ) = ( P 1 ) = 1 (mod q): 2 Lemma 3.21 Let (x 1 ; x 2 ; ; x 12 ) be a nontrivial monotone Boolean unction, invariant under the action o T 17. I is not elusive, then ( T 17 ) 1 (mod 2). Proo. Let H = h(1; 2; 3); (4; 5; 6); (7; 8; 9); (10; 11; 12)i: It is easy to veriy that H / T 17 and jhj = 3 4 : Moreover, jt 17 j = Thus, jt 17 =Hj = 2 5. Denote P = h(123)i. Clearly, P / H, jp j = 3, and jh=p j = 3 3. Furthermore, P is cyclic by Lemma By Corollary 2.1 and Lemma 3.21, ( T 17 ) 1 (mod 2). 2 Now we can prove our main theorem as ollows. Proo o Main Theorem. Let (x 1 ; x 2 ; ; x 12 ) be a nontrivial monotone weakly symmetric Boolean unction. By the denition o weakly symmetry, there exists a transitive permutation group G on 1; 2; ; 12g such that is invariant under G. By Lemma 3.1, G contains a transitive subgroup isomorphic to one o T i or i = 1; 2; ; 17, denoted by T. Note that T 1 is cyclic. When T = T i or i = 1; ; 14, by Lemma 2.4 and Lemmas , is elusive. When T = T i (i = 15; 16; 17), T has only one orbit on since it is transitive. This orbit cannot be a ace o. Otherwise, the monotonicity o orces being a constant, contradicting that is nontrivial. Thus, T = ;g and (T ) = 0. On the other hand, i is not elusive, then by Lemmas 3.17, 3.19, and 3.21, ( T ) 1 (mod p) where p = 2 or 3, a contradiction. 2 4 Discussion Rivest-Vuillemin conjecture is showed to be true or nontrivial monotone Boolean unctions o 12 variables. The proo involves many acts about permutation groups. With those acts, we established two new techniques which are used in dealing with groups T i or i = 9; ; 17. Reerences [1] I. Wegener, The Complexity o Boolean Functions, Wcley-Teubner Series in Comp. Sci., (New York{Stuttgart, 1987). [2] N. Nisan, CREW PRAM's and decision trees, SIAM J. Computing, 6 (1991)

18 [3] B. Bollobas, Extremal Graph Theory, (Academic Press, New York, 1978). [4] J.D. Dixon and B. Mortimer, Permutation Groups, (Springer, New York, 1996). [5] D.-Z. Du and K.-I Ko, Theory o Computational Complexity, (John-Wiley, New York, 2000) pp [6] S.-X. Gao, X.-D. Hu, and W. Wu, Nontrivial monotone weakly symmetric Boolean unctions o six variables are elusive, Theoretical Computer Science, 223 (1999) [7] S.-X. Gao, W. Wu, D.-Z. Du, and X.-D. Hu, Rivest-Vuillemin conjecture on monotone Boolean unctions is true or ten variables, Journal o Complexity, 15 (1999) [8] J. Greever, Stationary points or nite transormation groups, Duke Math. J., 27 (1960) [9] J. Kahn, M. Saks and D. Strutevant, A topological approach to evasiveness, Combinatorica, 4 (1984) [10] R. Oliver, Fixed-point sets o group actions on nite acyclic complexes, Comment. Math. Helvetici, 50 (1975) [11] R. L. Rivest and S. Vuillemin, A generalization and proo o the Aanderaa-Rosenberg conjecture, in Proc. o 7th ACM Symp. Theory o Computing, (Albuquerque, 1975) [12] J.J. Rotman, An Introduction to the Theory o Groups (4th edition) (Springer-Verlag, New York, 1994) [13] G.F. Royle, The transitive groups o degree twelve, J. Symbolic Computation, 4 (1987) [14] E.-F. Wang, Basic o Finite Group Theory, (Beijing University Press, 1987, in Chinese). 16